First part of $a$ comes from the Binomial Formula. Second part is done by multiplying with $(1+ \sqrt{2})^n$. Part $b$: If $n$ is even, take $m=a^2$ and so $m-1=2b^2$. Similar for $n$ odd. Seems to me like free $7$ points for the ones in the TST.

Part (a): We will induct on $n$. For $n=1$ it is trivial. Let when $1 \leq k \leq n$, there exist $a_k,b_k$ such that $(1-\sqrt{2})^k=a_k-\sqrt{2}b_k$ and $a_k^2-2b_k^2=(-1)^k$ $(1-\sqrt{2})^{n+1}=(a_n-b_n\sqrt{2})(1-\sqrt{2})=(a_n+2b_n)-(a_n+b_n)\sqrt{2}=a_{n+1}-b_{n+1}\sqrt{2}$ $(a_n+2b_n)^2-2(a_n+b_n)^2=-(a_n^2-2b_n^2)=(-1)^{n+1}$ So our induction is complete. Part (b): If $n$ is odd we take $m=2b^2$ and if $n$ is even we take $n=a^2$.